Find the polynomial function f with real coefficients that has the given degree, zeros, and solution point.

Question

The given zeros: -5, 1+sqrt3i

The given degree: 3

The given solution point: f(-2)=36

Answer 1

f(x) = k*(x+5) [( x +1) + i * sqrt(3)][ (x+1) - i*sqrt(3)]

= k * (x+5) [ (x+1)^2 - (-1)(3)]

= k ( x+5)[ (x+1)^2 + 3]

= k ( x+5)[ x^2 + 2x + 4]

f(-2) = 36 means

36 = k( -2+5)[ (-2)^2 + 2(-2) + 4]

= k (3) [ 4 - 4 + 4]

= 12k

k=3

f(x) = 3(x+5)(x^2 + 2x+4)